Given equation:

\( Ca(OH)_{2} + 2NH_{4}Cl \rightarrow CaCl_{2} + 2NH_{3} + 2H_{2}O . \)

Molar mass of \( NH_{4}Cl = 53.5 g/mole \)

Molar mass of \( CaCl_{2} = 111 g/mole \)

Molar mass of Ammonia \( (NH_{3}) = 17 g/mole \)

  1. The Mass Of Calcium Chloride Formed

Two moles of \( NH_{4}Cl react to give one mole of CaCl_{2} \)

Two moles of \( NH_{4}Cl will give \) \( \Rightarrow \frac{0.1 * 1}{2} \) \( \Rightarrow 0.05 moles of CaCl_{2} \) \( Mass of CaCl_{2} = 0.05 * 111 \) \( \Rightarrow 5.55 grams of CaCl_{2} \)

Therefore,the mass of Calcium Chloride formed is 5.55 gms

  1. The Volume, At S.T.P Of Ammonia Liberated.

Two moles of \( NH_{4}Cl gives one mole of NH_{3} gas \) \( 0.1 moles of NH_{4}Cl will react to give \) \( \Rightarrow \frac{0.1}{2} = 0.05 moles of NH_{3} gas \)

One mole of Ammonia \( (NH_{3}) at STP = 2.4 L \) \( 0.05 mole of Ammonia (NH_{3}) at STP = 22.4 * 0.5 \) \( \Rightarrow 1.12 L of NH_{3} gas \)

Therefore, the volume, at S.T.P of Ammonia gas liberated is 1.12 L

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